Question

CH\equiv \equiv CH\xrightarrow[{{H}_{2}}S{{O}_{4}}]{{{H}_{2}}O/H{{g}^{2+}}}X\xrightarrow{LiAl{{H}_{4}}}Y\xrightarrow{{{P}_{4}}/B{{r}_{2}}}Z, Here Z is:

• A. ethylene bromide
• B. ethanol
• C. ethyl bromide
• D. ethylidene bromide

Answer 1

Answer: (C)

ethyl bromide

Answer 2

The reaction is as follows CH\equiv CH\xrightarrow[{{H}_{2}}S{{O}_{4}}]{{{H}_{2}}O/H{{g}^{2+}}}C{{H}_{3}}-CHO $\xrightarrow[(reduction)]{LiAl{{H}_{4}}}C{{H}_{3}}C{{H}_{2}}OH\xrightarrow{{{P}_{4}}/B{{r}_{2}}}\underset{Ethyl\,bromide}{\mathop{C{{H}_{3}}C{{H}_{2}}Br}}\,$